已知等差数列｛An｝的前三项为a-1,4,2a,记前n项和为Sn.（1）,设Sk等于2550,求a和k的值,（2）,设bn等于n分之Sn,求b3+b7+b11+……+b4n-1的值

题目

已知等差数列｛An｝的前三项为a-1,4,2a,记前n项和为Sn.（1）,设Sk等于2550,求a和k的值,（2）,设bn等于n分之Sn,求b3+b7+b11+……+b4n-1的值

答案

(1)
an=a1+(n-1)d
2a+(a-1) = 8
a=3
a1= a1-1=2
d= 4-(a-1)= 2
an = 2n
Sn =n(n+1)
Sk=k(k+1)=2550
k^2+k -2550=0
(k+51)(k-50)=0
k=50
(2)
bn = Sn/n
= n+1
b3+b7+b11+...+b(4n-1)
=4+8+12+.+4n
=4(1+2+...+n)
=2n(n+1)